Problem Solving

Night reader wrote:How many four-digit odd numbers do not use any digit more than once?
(A) 1728
(B) 2160
(C) 2240
(D) 2268
(E) 2520

Let the number be ABCD.

D can be filled in 5 ways (1,3,5,7,9), as the number is odd.
A can be filled in 8 ways (1-9 but not the ones used in D; First digit can't be zero)
B can be filled in 8 ways (0-9 but not the ones used in C and D)
C can be filled in 7 ways (0-9 but not the ones used in B, C and D)

Thus total number of such numbers = 5*8*8*7 = 2240

The correct answer is C.

swathi8388 wrote:Though i understood the answer I am confused why I wont get the Ssame answer the other way..that is..find the total number 4 digit numbers that have unique digits and divide it by 2 (cause in any set of numbers the odd and even numbers have to be equal?)

but in that case we get 2268 :

9 * 9 * 8 * 7 = 648 * 7 = 4536

4536 / 2 = 2268

can someone explain.

You are assuming that the 4536 numbers are such that half are odd and half are even. This is not the case.
Since we are, in essence, saying that the thousands digit cannot be zero (an even number), then we are increasing the likelihood that zero appears among the other 3 digits (including the units digit).
So, among those 4536 numbers, more than half are even, which means fewer than half are odd.
So, when you divided 4536 by 2, the result is too big.

Cheers,
Brent

Night reader wrote:How many four-digit odd numbers do not use any digit more than once?

(A) 1728
(B) 2160
(C) 2240
(D) 2268
(E) 2520

Rahul's solution above is great. It applies something called the By the Fundamental Counting Principle (FCP). This technique is also called the Slot Method.

It's important to note, that when you apply this strategy, you should always begin with the most restrictive cases

Here's what I mean.

We're going to take the task of building acceptable 4-digit numbers and break it into stages.
Our stages will consist of selecting the four digits needed. Among these stages, the most restrictive one is the units digit, since it must be odd. So, we'll begin there.

Stage 1: Select the units digit
Since the digit must be 1, 3, 5, 7, or 9, we can complete this stage in 5 ways.

From here, the 2nd most restrictive stage is selecting the thousands digit (since it cannot be zero). So . ..

Stage 2: Select the thousands digit
This digit cannot equal zero and it cannot be the same as the digit selected as the units digit.
This leaves 8 digits to choose from, so we can complete this stage in 8 ways.

At this point, the two remaining stages are equally restrictive, so we can complete either stage next.

Stage 3: Select the tens digit
This digit cannot equal either of the 2 digits already-selected digits.
This leaves 8 digits to choose from, so we can complete this stage in 8 ways.

Stage 4: Select the hundreds digit
This digit cannot equal either of the 3 digits already-selected digits.
This leaves 7 digits to choose from, so we can complete this stage in 7 ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus build a 4-digit integer) in (5)(8)(8)(7) ways ([spoiler]= 2240 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775